Question 1

Consider a lossless antenna with a directive gain of +6 dB. If 1 mW of input power is fed to the antenna, what will be the total power radiated by the antenna?

Accepted Answer

Introduction / Context:Antennas are described using different gain measures, such as directive gain, power gain, and efficiency. This question examines the difference between directive gain and radiated power, focusing on the principle that directive gain affects the directional distribution of radiated energy but not the total radiated power for a lossless antenna.Given Data / Assumptions:Input power to antenna = 1 mW.Directive gain = +6 dB.Antenna is lossless (no ohmic or dielectric losses).Concept / Approach:For a lossless antenna, all input power is radiated. Directive gain (in dB) indicates how much stronger the field or power density is in the maximum radiation direction compared to an isotropic radiator. It does not change the total radiated power, only its spatial distribution. Therefore, the total radiated power always equals the input power provided losses are negligible.Step-by-Step Solution:Step 1: Convert gain from dB to linear scale. Gd = 10^(6/10) = 10^0.6 ≈ 3.98.Step 2: Interpret directive gain: maximum radiation intensity is 3.98 times that of an isotropic radiator for same power.Step 3: For a lossless antenna, Pin = Prad. Here, Pin = 1 mW.Step 4: Thus Prad = 1 mW.Step 5: Gain affects directivity but does not alter total radiated power.Verification / Alternative check:If antenna efficiency = 100%, input power and radiated power are identical regardless of gain. Directive gain only shapes radiation pattern.Why Other Options Are Wrong:4 mW: Misinterpretation, assuming gain multiplies power rather than intensity.7 mW: No basis in gain-power relation.1/4 mW: Would imply efficiency 0.5 mW: Suggests 50% efficiency, not given here.Common Pitfalls:Confusing directive gain with power gain; assuming that dB gain changes total radiated power; ignoring lossless assumption.Final Answer:1 mW

Question 2

What is the radiation resistance of a short dipole antenna with length equal to λ/10?

Accepted Answer

Introduction / Context:Radiation resistance is the equivalent resistance of an antenna that represents the power radiated as electromagnetic waves. Short dipoles (length ≪ λ) have radiation resistance that depends on their electrical length. This problem requires applying the standard formula for a dipole with length λ/10.Given Data / Assumptions:Antenna length L = λ/10.Type: short dipole (length much less than half a wavelength).Free-space radiation resistance formula is applicable.Concept / Approach:For a short dipole antenna: Rr ≈ 80π^2 (L/λ)^2. Substituting L/λ = 1/10 gives a numerical value.Step-by-Step Solution:Step 1: Formula: Rr = 80π^2 (L/λ)^2.Step 2: Substitute L/λ = 1/10 = 0.1.Step 3: Compute (0.1)^2 = 0.01.Step 4: Multiply: 80 * π^2 * 0.01 ≈ 80 * 9.8696 * 0.01 ≈ 7.9 Ω.Step 5: Closest option is 7.8 Ω.Verification / Alternative check:Cross-check with known rule: Radiation resistance for λ/2 dipole ≈ 73 Ω; for λ/10 dipole it should be much smaller (~8 Ω), consistent with result.Why Other Options Are Wrong:8 Ω: Approximation but less precise than 7.8 Ω.780 Ω and 78 Ω: Orders of magnitude too high.1 Ω: Too low compared with formula prediction.Common Pitfalls:Forgetting the squared term (L/λ)^2; using λ/2 dipole values; confusing input impedance with radiation resistance.Final Answer:7.8 Ω

Question 3

A transmission line feeds 1 watt of power to a horn antenna with a gain of 10 dB. The antenna is matched to the transmission line. What is the total power radiated into free space?

Accepted Answer

Introduction / Context:Horn antennas are directive antennas often characterized by high gain values in dB. This question clarifies the relationship between input power, gain, and total radiated power, highlighting that gain shapes directivity but does not amplify the actual radiated power beyond the input for a matched, lossless antenna.Given Data / Assumptions:Input power Pin = 1 W.Antenna gain = 10 dB.Matched system, no reflections or mismatches.Concept / Approach:Power gain (in dB) indicates how power is concentrated in certain directions relative to isotropic radiation, not an increase in total power. For a lossless, matched antenna, Prad = Pin. The 10 dB gain indicates 10× more power density in the main lobe direction than an isotropic radiator of equal total power.Step-by-Step Solution:Step 1: Convert 10 dB gain to linear: G = 10^(10/10) = 10.Step 2: Interpret: Maximum directional intensity is 10× isotropic, but integrated power over all space remains equal to Pin.Step 3: Input Pin = 1 W. Lossless antenna → Prad = Pin = 1 W.Step 4: Thus total power radiated is 1 W.Verification / Alternative check:Compare with directive gain example: Same principle as earlier; gain affects pattern, not total power.Why Other Options Are Wrong:10 W: Misinterpretation that gain multiplies total power.0.1 W or 0.01 W: Suggest reduced power, contradicting matched, lossless assumption.5 W: Arbitrary scaling; no basis in antenna theory.Common Pitfalls:Confusing antenna gain with amplifier gain; assuming gain increases actual power rather than concentrating it.Final Answer:1 watt

Question 4

What is the approximate radiation resistance of a λ/16 wire dipole antenna in free space?

Accepted Answer

Introduction / Context:Radiation resistance is the effective resistance representing power radiated as electromagnetic energy. For dipoles shorter than half a wavelength, the radiation resistance decreases sharply with decreasing length. This question asks for the resistance of a dipole λ/16 long.Given Data / Assumptions:Dipole length = λ/16.Operating in free space.Standard approximation formula applies.Concept / Approach:The approximate formula for short dipoles: Rr ≈ 80π^2 (L/λ)^2. For a dipole, when length = λ/16, we substitute to calculate the value.Step-by-Step Solution:Step 1: Length ratio: L/λ = 1/16 = 0.0625.Step 2: Square it: (0.0625)^2 = 0.00390625.Step 3: Multiply constant: 80π^2 ≈ 789.57.Step 4: Compute: 789.57 * 0.00390625 ≈ 3.08 Ω.Step 5: Therefore Rr ≈ 3 Ω.Verification / Alternative check:Known reference: λ/2 dipole ≈ 73 Ω, λ/16 dipole ≈ few ohms; result is consistent.Why Other Options Are Wrong:1 Ω: Underestimates resistance; requires even shorter length.13 Ω and 30 Ω: Overestimates by factors of 4–10.7.3 Ω: Closer to λ/10 dipole, not λ/16.Common Pitfalls:Applying half-wave dipole resistance values to short dipoles; ignoring quadratic dependence on (L/λ).Final Answer:3 Ω

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